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| 1. |
convert -3i into polar form. |
| Answer» Let\xa0{tex}-3i= 0-3i=r\\left( cos\\theta +isin\\theta \\right) {/tex}Comparing real and imaginary parts{tex}rcos\\theta =0.......................\\left( i \\right) \\\\ rsin\\theta =-3....................\\left( ii \\right) \\\\ {/tex}Squaring and adding equations (i) and (ii){tex}{ r }^{ 2 }{ cos }^{ 2 }\\theta +{ { r }^{ 2 } }sin^{ 2 }\\theta =9\\\\ \\Rightarrow { r }^{ 2 }=9\\quad \\quad \\quad \\quad \\quad \\quad \\quad \\left( { \\because cos }^{ 2 }\\theta +sin^{ 2 }\\theta =1 \\right) {/tex}{tex}\\Rightarrow r=3{/tex}Substituting\xa0{tex}r=3{/tex}\xa0in equations (i) and (ii), we get\xa0{tex}cos\\theta =0\\quad and\\quad sin\\theta =-1\\\\ {/tex}So\xa0{tex}\\theta {/tex}\xa0lies in fourth quadrant\xa0{tex}\\therefore \\quad \\theta =\\frac { -\\Pi }{ 2 } {/tex} [ in the fourth quarant format of amplitude or principal argument is\xa0{tex}-\\theta {/tex}\xa0(where {tex}\\theta =\\frac { \\Pi }{ 2 } {/tex}\xa0)\xa0]{tex}-3i=3\\left( cos\\left( \\frac { -\\Pi }{ 2 } \\right) +isin\\left( \\frac { -\\Pi }{ 2 } \\right) \\right) \\\\ \\Rightarrow -3i=3\\left( cos\\left( \\frac { \\Pi }{ 2 } \\right) -isin\\left( \\frac { \\Pi }{ 2 } \\right) \\right) \\quad \\quad \\left[ \\because cos\\left( -\\theta \\right) =cos\\theta \\quad ,sin\\left( -\\theta \\right) =-sin\\theta \\right] {/tex}\xa0 | |